Question

For a reaction,
$N_2(g)+3H_2(g)\to 2N{H}_3(g)$; identify dihydrogen $(H_2)$ as a limiting reagent in the following reaction mixtures.

• A. $14g$ of $N_2$ + $4g$ of $H_2$
• B. $28g$ of $N_2$ + $6g$ of $H_2$
• C. $56g$ of $N_2$ + $10g$ of $H_2$
• D. $35g$ of $N_2$ + $8g$ of $H_2$

Answer 1

Answer: (C)

When $56g$ of $N_2+10g$ of $H_2$ is taken as a combination then dihydrogen $\left(H_2\right)$ act as a limiting reagent in the reaction.
...(I)
$28g{N}_2$ requires $6g{H}_2$ gas.
$56g$ of $N_2$ requires $\frac{6g}{28g}\times 56g$
$=12g$ of $H_2$
$12g$ of $H_2$ gas is required for $56g$ of $N_2$ gas but only $10g$ of $H_2$ gas is present in option (a).
Hence, $H_2$ gas is the limiting reagent.
In option (b), i.e. $35g$ of $N_2+8g$ of $H_2$.
As $28g{N}_2$ required $6g$ of $H_2$.
$35g{N}_2$ required $\frac{6g}{28g}\times 35g{H}_2$
$\Rightarrow 7.5g$ of $H_2$.
Here, $H_2$ gas does not act as limiting reagent since $7.5g$ of $H_2$ gas is required for $35g$ of $N_2$ and $8g$ of $H_2$ is present in reaction mixture.
Mass of $H_2$ left unreacted $=8-7.5g$ of $H_2=0.5g$ of $H_2$
Similarly, in option (c) and (d), $H_2$ does not act as limiting reagent.
For $14g$ of $N_2+4g$ or $H_2$.
As we know, $28g$ of $N_2$ reacts with $6g$ of $H_2$.
$14g$ of $N_2$ reacts with $\frac{6}{28}\times 14g$ of $H_2$
$\Rightarrow 3g$ of $H_2$.
For $28g$ of $N_2+6g$ or $H_2$,
i.e. $28g$ of $N_2$ reacts with $6g$ of $H_2$ (by equation $I$ ).